Problem: You have found the following ages (in years) of all 4 zebras at your local zoo: $ 3,\enspace 23,\enspace 8,\enspace 5$ What is the average age of the zebras at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 4 zebras at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{3 + 23 + 8 + 5}{{4}} = {9.8\text{ years old}} $ Find the squared deviations from the mean for each zebra. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $3$ years $-6.8$ years $46.24$ years $^2$ $23$ years $13.2$ years $174.24$ years $^2$ $8$ years $-1.8$ years $3.24$ years $^2$ $5$ years $-4.8$ years $23.04$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{46.24} + {174.24} + {3.24} + {23.04}} {{4}} $ $ {\sigma^2} = \dfrac{{246.76}}{{4}} = {61.69\text{ years}^2} $ The average zebra at the zoo is 9.8 years old. The population variance is 61.69 years $^2$.